GSEB Class 10 Maths Solutions Chapter 1 Real Numbers Exercise 1.4

 

Question 1. Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion.
1. \( \frac{13}{3125} \)
2. \( \frac{17}{8} \)
3. \( \frac{64}{455} \)
4. \( \frac{15}{1600} \)
5. \( \frac{29}{343} \)
6. \( \frac{23}{2^{3} \times 5^{2}} \)
7. \( \frac{129}{2^{2} \times 5^{7} \times 7^{5}} \)
8. \( \frac{6}{15} \)
9. \( \frac{35}{50} \)
10. \( \frac{77}{210} \)
Answer:
A rational number \( \frac{p}{q} \) (in its simplest form) has a terminating decimal expansion if the prime factorization of its denominator \( q \) is of the form \( 2^n \times 5^m \), where \( n \) and \( m \) are non-negative integers. Otherwise, it will have a non-terminating repeating decimal expansion.

(i) \( \frac{13}{3125} \)

The denominator \( 3125 = 5^5 \). This matches the form \( 2^n \times 5^m \) (where \( n=0 \) and \( m=5 \)). Hence, the decimal expansion of \( \frac{13}{3125} \) is terminating.
In simple words: The bottom number, 3125, is only made of fives when you break it down. This means the decimal will stop, it won't go on forever.

(ii) \( \frac{17}{8} \) We can write \( 8 = 2^3 \). The denominator matches the form \( 2^n \times 5^m \) (where \( n=3 \) and \( m=0 \)). Hence, the decimal expansion of \( \frac{17}{8} \) is terminating.
In simple words: The bottom number, 8, is only made of twos. This means the decimal expansion will eventually end.

(iii) \( \frac{64}{455} \)

We can write \( 455 = 5^1 \times 7^1 \times 13^1 \). Since the denominator is not in the form \( 2^n \times 5^m \) and it also contains 7 and 13 as factors, its decimal expansion will be non-terminating repeating.
In simple words: The bottom number, 455, has sevens and thirteens in it, not just twos and fives. So, the decimal will go on forever with a pattern that repeats.

(iv) \( \frac{15}{1600} \)

We simplify the fraction: \( \frac{15}{1600} = \frac{3 \times 5}{320 \times 5} = \frac{3}{320} \). The denominator \( 320 = 2^6 \times 5^1 \). This matches the form \( 2^n \times 5^m \). Hence, the decimal expansion of \( \frac{15}{1600} \) is terminating.
In simple words: After simplifying, the bottom number 320 is only made of twos and fives. So, this decimal will stop.

(v) \( \frac{29}{343} \) We can write \( 343 = 7^3 \). Since the denominator is not in the form \( 2^n \times 5^m \) and it also has 7 as a factor, its decimal expansion will be non-terminating repeating.
In simple words: The bottom number, 343, is made of sevens. This means the decimal will go on forever and repeat in a pattern.

(vi) \( \frac{23}{2^{3} \times 5^{2}} \) The denominator is already in the form \( 2^n \times 5^m \) (where \( n=3 \) and \( m=2 \)). Hence, the decimal expansion is terminating.
In simple words: The bottom part of the fraction is already in the right form (only twos and fives). So, this decimal will stop.

(vii) \( \frac{129}{2^{2} \times 5^{7} \times 7^{5}} \) Since the denominator has 7 as a factor, it is not in the form \( 2^n \times 5^m \). Hence, the decimal expansion of \( \frac{129}{2^{2} \times 5^{7} \times 7^{5}} \) is non-terminating repeating.
In simple words: The bottom part has a seven in it, which is not a two or a five. This means the decimal will go on forever with a repeating pattern.

(viii) \( \frac{6}{15} \) We simplify the fraction: \( \frac{6}{15} = \frac{2 \times 3}{5 \times 3} = \frac{2}{5} \). The denominator \( 5 = 2^0 \times 5^1 \). This matches the form \( 2^n \times 5^m \). Hence, the decimal expansion of \( \frac{6}{15} \) is terminating.
In simple words: When you simplify the fraction, the bottom number becomes 5. Since 5 is a power of 5, the decimal will stop.

(ix) \( \frac{35}{50} \) We simplify the fraction: \( \frac{35}{50} = \frac{5 \times 7}{5 \times 10} = \frac{7}{10} = \frac{7}{2 \times 5} \). The denominator \( 2 \times 5 \) is in the form \( 2^n \times 5^m \). Hence, the decimal expansion of \( \frac{35}{50} \) is terminating.
In simple words: When you simplify the fraction, the bottom number becomes 10 (which is 2 times 5). Since it's made of only twos and fives, the decimal will stop.

(x) \( \frac{77}{210} \) We simplify the fraction: \( \frac{77}{210} = \frac{11 \times 7}{3 \times 7 \times 2 \times 5} = \frac{11}{3 \times 2 \times 5} \). Since the denominator has 3 as a factor, it is not in the form \( 2^n \times 5^m \). Hence, the decimal expansion is non-terminating.
In simple words: After simplifying, the bottom number 30 has a three in it, not just twos and fives. This means the decimal will go on forever with a repeating pattern.

Exam Tip: To quickly determine if a decimal expansion terminates, always simplify the fraction first, then prime factorize the denominator. If the prime factors are only 2s and 5s, it terminates.

 

Question 2. Write down decimal expansions of the rational numbers in question 1 above which have terminating decimal expansions.
Answer:
(i) \( \frac{13}{3125} \)
We have \( \frac{13}{3125} = \frac{13}{5^5} = \frac{13 \times 2^5}{5^5 \times 2^5} = \frac{13 \times 32}{(5 \times 2)^5} = \frac{416}{10^5} = \frac{416}{100000} = 0.00416 \).

(ii) \( \frac{17}{8} \)
We have \( \frac{17}{8} = \frac{17}{2^3} = \frac{17 \times 5^3}{2^3 \times 5^3} = \frac{17 \times 125}{(2 \times 5)^3} = \frac{2125}{10^3} = \frac{2125}{1000} = 2.125 \).

(iii) \( \frac{15}{1600} \)
We have \( \frac{15}{1600} = \frac{3 \times 5}{320 \times 5} = \frac{3}{320} \). Now, \( 320 = 2^6 \times 5^1 \). So, \( \frac{3}{2^6 \times 5^1} = \frac{3 \times 5^5}{2^6 \times 5^6} = \frac{3 \times 3125}{(2 \times 5)^6} = \frac{9375}{10^6} = \frac{9375}{1000000} = 0.009375 \).
(iv) \( \frac{23}{2^{3} \times 5^{2}} \)
We have \( \frac{23}{2^{3} \times 5^{2}} = \frac{23 \times 5}{2^{3} \times 5^{2} \times 5} = \frac{115}{2^{3} \times 5^{3}} = \frac{115}{(2 \times 5)^3} = \frac{115}{10^3} = \frac{115}{1000} = 0.115 \).

(v) \( \frac{6}{15} \)
We have \( \frac{6}{15} = \frac{2 \times 3}{5 \times 3} = \frac{2}{5} = \frac{2 \times 2}{5 \times 2} = \frac{4}{10} = 0.4 \).

(vi) \( \frac{35}{50} \)
We have \( \frac{35}{50} = \frac{5 \times 7}{5 \times 10} = \frac{7}{10} = 0.7 \).
In simple words: For each fraction that stops, we turn it into a decimal by making the bottom number a power of 10. Then, we simply write the top number with the decimal point in the right place.

Exam Tip: To convert a fraction with a denominator of the form \( 2^n \times 5^m \) into a decimal, multiply the numerator and denominator by the appropriate power of 2 or 5 to make the denominator a power of 10 (e.g., \( 10^k \)).

 

Question 3. The following real numbers have decimal expansions as given below. In each case decide whether they are rational or not. If they are rational and of the form \( \frac{p}{q} \), what can you say about the prime factor of \( q \)?
1. 43.123456789
2. 0.120 1200 12000 120000 ...
3. \( 43.1\overline {23456789} \)
Answer:
1. The number 43.123456789 has a terminating decimal expansion.
\( 43.123456789 = \frac{43123456789}{10^9} \)
Since this number has a terminating decimal expansion, it is a rational number. If it is expressed in the form \( \frac{p}{q} \), the prime factors of \( q \) will only be 2 and 5.
In simple words: This number stops after a few digits, so it's a rational number. If we write it as a fraction, the bottom part of the fraction will only have 2s and 5s as its building blocks.

2. The number 0.120 1200 12000 120000 ... has a decimal expansion that is neither terminating nor repeating. Therefore, the given number is an irrational number.
In simple words: This number goes on forever without any pattern that repeats. So, it's an irrational number, which means it can't be written as a simple fraction.

3. The number \( 43.1\overline {23456789} \) has a decimal expansion that is non-terminating but repeating. Since the decimal expansion is non-terminating but recurring, the given number is a rational number. If it is expressed in the form \( \frac{p}{q} \), the prime factors of \( q \) will have factors other than 2 or 5.
In simple words: This number goes on forever but it has a repeating pattern. So, it's a rational number. If we write it as a fraction, the bottom part will have other prime numbers (like 3 or 7) in its building blocks, not just 2s and 5s.

Exam Tip: Remember that rational numbers either terminate or repeat, while irrational numbers are non-terminating and non-repeating. The prime factors of the denominator for a rational number indicate its decimal type.